Is ${995754}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {995754}= &&{9}\cdot100000+ \\&&{9}\cdot10000+ \\&&{5}\cdot1000+ \\&&{7}\cdot100+ \\&&{5}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {995754}= &&{9}(99999+1)+ \\&&{9}(9999+1)+ \\&&{5}(999+1)+ \\&&{7}(99+1)+ \\&&{5}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {995754}= &&\gray{9\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {9}+{9}+{5}+{7}+{5}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${995754}$ is divisible by $3$ if ${ 9}+{9}+{5}+{7}+{5}+{4}$ is divisible by $3$ Add the digits of ${995754}$ $ {9}+{9}+{5}+{7}+{5}+{4} = {39} $ If ${39}$ is divisible by $3$ , then ${995754}$ must also be divisible by $3$ Add the digits of ${39}$ $ {3}+{9} = \color{#9D38BD}{12} $ If $\color{#9D38BD}{12}$ is divisible by $3$ , then ${39}$ must also be divisible by $3$ $\color{#9D38BD}{12}$ is divisible by $3$, therefore ${995754}$ must also be divisible by $3$.